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Question
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Solution
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol−1
∴ Number of moles present in 1000 g of water = `1000/18`
= 55.56 mol
Therefore, the mole fraction of the solute in the solution is
`"x"_2 = 1/(1+55.56) = 0.0177`
Vapour pressure of water, `"p"_1^0` = 12.3 kPa
Applying the relation, `("p"_1^0-"p"_1)/"p"_1^0 = "x"_2`
`=> (12.3- "p"_1)/12.3 = 0.0177`
`=>12.3 - "p"_1 = 0.2177`
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
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